$${n_{Si} = \frac{(2330\, \mathrm{kg/m^3})(6,02 \times 10^{23}\, \mathrm{aatomit/mol})}{0,0281\, \mathrm{kg/mol}} \\ = 5, \times 10^{28}\, \mathrm{aatomit/m^3} = 5 x 10^{29}\, \mathrm{m^3}}$$